Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 3x}{x + 6} = \dfrac{-6x + 18}{x + 6}$
Solution: Multiply both sides by $x + 6$ $ \dfrac{x^2 - 3x}{x + 6} (x + 6) = \dfrac{-6x + 18}{x + 6} (x + 6)$ $ x^2 - 3x = -6x + 18$ Subtract $-6x + 18$ from both sides: $ x^2 - 3x - (-6x + 18) = -6x + 18 - (-6x + 18)$ $ x^2 - 3x + 6x - 18 = 0$ $ x^2 + 3x - 18 = 0$ Factor the expression: $ (x + 6)(x - 3) = 0$ Therefore $x = -6$ or $x = 3$ At $x = -6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -6$, it is an extraneous solution.